A person is making iced tea by brewing hot tea and adding ice untilthe tea cools

Question

A person is making iced tea by brewing hot tea and adding ice untilthe tea cools. How much ice, at an initial temperature of-10oC, should be added to a 2.00x10-4m3 glass of tea at 95.0oC to cool the tea to100C? Assume the glass has negligible heat stored.

Explanation / Answer

Process : Ice of -10 degree will come to ice at zero degree andthen take latent heat to come at water at zero degree then thetemperature of this water will rise to final temperature i.e 10degree. On the other hand, Tea at 95 degree will lose heat to come at 10degree Let tea as a form of Hot water Volume of tea = 2.00x10-4 m3 => mass of tea = 2.00x10-4 * 1000 = 0.2 kg =200 g (density of water = 1000 kg/m3) heat loss by tea, Q = 200*4.186*(95-10) = 71162 J Heat gain by ice ,Q' = mSiT + mL + mSwT where, m = mass of ice Si = specifi heat of ice = 2.06 J/gC Sw = specific heat of water = 4.186 J/gC L = latent heat = 333.55 J/g => Q' = m*2.06*(0-(-10)) + m*333.55 + m*4.186*(10-0) = 20.6m +333.55m + 41.86m = 396.01m By Principle of calorimetry HEAT LOSS =HEAT GAIN 71162 = 396.01m => m = 179.697 g Hence mass od ice = 179.697 grams

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